MCQ
$\cos^230^\circ\cot^245^\circ+4\sec^260^\circ+\frac{1}{2}\cos^290^\circ-2\tan^260^\circ=?$
- A$\frac{73}{8}$
- B$\frac{75}{8}$
- C$\frac{81}{8}$
- ✓$\frac{83}{8}$
✓
Answer
Correct option: D.
$\frac{83}{8}$
$\cos^230^\circ\cot^245^\circ+4\sec^260^\circ+\frac{1}{2}\cos^290^\circ-2\tan^260^\circ$
$=\Big(\frac{\sqrt3}{2}\Big)^2\times\Big(\frac{1}{\sqrt2}\Big)^2+4(2)^2+\frac{1}{2}\times0-2\big(\sqrt3\big)^2$
$=\frac{3}{4}\times\frac{1}{2}+16+0-6$
$=\frac{3}{8}+10$
$=\frac{3+80}{8}$
$=\frac{83}{8}$
$=\Big(\frac{\sqrt3}{2}\Big)^2\times\Big(\frac{1}{\sqrt2}\Big)^2+4(2)^2+\frac{1}{2}\times0-2\big(\sqrt3\big)^2$
$=\frac{3}{4}\times\frac{1}{2}+16+0-6$
$=\frac{3}{8}+10$
$=\frac{3+80}{8}$
$=\frac{83}{8}$
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